Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
x = -8
y = 12

I was told to calculate the rate of x according to y, and viceversa. So what I did was divide them.

x/y = -0.66
y/x = -1.5

My question is rather straightforward: Is this right?

I am just unsure if there can be negative rates. For instance, I though that since y is greater than x, y I would get a positive rate.

share|improve this question
1  
what do you mean by "rate"? –  Aang Sep 25 '12 at 1:49
    
@Avatar: Sorry, actually I don't speak English natively (Spanish). My word is "tasa" and apparently it translates to "rate" or "valuation". It is usually applied to compare the impact of two things, as in "Comparing these two quakes, the rate is 1:3, meaning that one quake was three times greater than the other" –  Zol Tun Kul Sep 25 '12 at 1:54
    
If one of the two numbers is negative, then "rate" would surely be negative. –  Aang Sep 25 '12 at 2:02
    
Oh, so... plain division is alright? Well, that's.. That's good. Yeah. Thanks.. answered. –  Zol Tun Kul Sep 25 '12 at 2:04
1  
@Avatar I think the English word you are looking for is ratio. –  Robert Miller Sep 25 '12 at 2:10

1 Answer 1

up vote 1 down vote accepted

When you multiply by something negative, the product is on the other side of 0 from where you started. Similarly when you divide two numbers, the quotient is positive if both are positive or both are negative. Are you happy with the fact that the product of two negative numbers is positive?

share|improve this answer
    
An interesting side note: When the housing market started to crash, a number of firms' programs failed because they assumed that housing prices would always go up. Believe it or not!!! –  marty cohen Sep 25 '12 at 6:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.